TÌM X BIẾT:
1/10+1/40+1/88+...+3/(X+2)x(3X+5)
Tìm x biết 1/10 + 1/40 + 1/88 + ... + 1/(3x + 2)(3x + 5) = 4/25.
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
Tìm x thuộc Z, biết
1/10 + 1/40 + 1/88 + ... + 1/(3x+2)(3x+5)
giải nhanh + đúng + đầy đủ + kiểu tường thuật mình tích
ta có : 1/10 = 1/2.5
1/40 = 1/5.8
suy ra 1/10 + 1/40 + 1/88 + ... + 1/(3x+2).(3x+5) = 3/20 = 1/2.5 + 1/5.8 + 1/8.11 +....+ 1/(3x+2).(3x+5)
= 1/3(1/2 - 1/5) + 1/3(1/5 - 1/8) + ...+ 1/3(1/8 - 1/11) + ....+ 1/3 [(3x+2) (3x+5)] = 3/20
= 1/3(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +.... + 1/3x+2 + 1/3x+5) = 3/20
1/3 ( 1/2 - 1/3x+5) = 3/20
suy ra 1/2 - 1/3x+5 = 3/20 : 1/3 = 9/20
suy ra 1/3x+5 = 1/2 - 9/20 = 1/20
suy ra 3x+5 = 20
suy ra 3x = 20 - 5= 15
suy ra x = 15 : 3 = 5
vay x = 5
nhớ k nha!!!!
Tìm x thuộc Z sao cho 1/10+1/40+1/88+...+1/(3x+2).(3x+5)+ 4/25
1) tìm x thuộc Z sao cho ( 3x+4) chia hết cho (x-1)
2) tìm x biết (15/10 - 15/40 + 15/88 + 30/11.17 + 15/17.20) . x = 3/4
Tìm x biết:
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{3}{\left(x+2\right)x\left(3x+5\right)}=\frac{3}{20}\)
giải giúp tôi nhé!arigator
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
Bài 1
a) 5 . | x - 3 | = \(\frac{5}{7}\)
b) (\(\frac{2}{3}\)x - \(\frac{1}{3}\)) . ( \(\frac{4}{5}\)x + \(\frac{2}{3}\)) = 0
Bài 2: Tìm x \(\varepsilon\) Z, biết
1/10 + 1/40 + + 1/88 + ... + 1/(3x+2).(3x+5)
Giải đúng + nhanh + có lời giải tường thuật mình tích cho
Tìm x
a/\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}+x\div\frac{1}{3}=-4\)
b/\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right)\left(3x+5\right)}=\frac{3}{20}\)
1/10+1/40+1/88+...+1/(x+2)(x+5)=3/20
làm lại
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{20}\)
=>\(x+5=20\)
=>\(x=20-5\)
=>\(x=15\)
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)
=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{20}\)
=>\(x+3=20\)
=>\(x=20-3\)
=>\(x=17\)
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x+2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}:\frac{1}{3}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}\cdot3\\ \frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{1}{2}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{10}{20}\\ \frac{1}{x+5}=\frac{19}{20}\\ \frac{19}{19\left(x+5\right)}=\frac{19}{20}\\ \Rightarrow19\left(x+5\right)=20\\ 19x+95=20\\ 19x=20-95\\ 19x=-75\\ x=\left(-75\right):19\\ x=\frac{75}{19}\)
Vậy \(x=\frac{75}{19}\)